Determinant and Linearly Independent Vectors

Question

Question: Let $A\in M_{3\times3}$ and $x,y,z\in\mathbb R^3$. If $x,y,z$ are linearly independent and we have that,

$Ax = \begin{bmatrix} 1\\0\\1 \end{bmatrix}, Ay=\begin{bmatrix} 0\\1\\0 \end{bmatrix},Az=\begin{bmatrix} 1\\1\\1 \end{bmatrix}$

Find the determinant of A.

I noticed that $Ax+Ay=Az$, but I am not sure of what this tells me about A to help me find its determinant.

Answer 1

Hint:

If $A$ is invertible, it maps a basis onto a basis.

Answer 2

The fact that $Ax+Ay=Az$ tells you that the set $\{Ax,Ay,Az\}$ is not linearly independent. Therefore, that determinant that you're after is equal to $0$.

Answer 3

$ A=\left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \\ \end{array} \right)$
where entries are in $\mathbb{R}$
Linear Independence
$a_1 \text{Σ}\left[x_i\right]+a_2 \text{ Σ}\left[y_i\right]+a_3 \text{ Σ}\left[z_i\right]=0$ -----(A)
$\Rightarrow a_1=a_2=a_3=0$
$a_1,a_2,a_3 \in \mathbb{R}$
$\text{Ax}+\text{Ay}+\text{Az}=\left( \begin{array}{c} 2 \\ 2 \\ 2 \\ \end{array} \right)$
$a \text{Σ}\left[x_i\right]+b \text{ Σ}\left[y_i\right]+c \text{ Σ}\left[z_i\right]=0$ -----(B)
$d \text{Σ}\left[x_i\right]+e \text{ Σ}\left[y_i\right]+f \text{ Σ}\left[z_i\right]=0$
$g \text{Σ}\left[x_i\right]+h \text{ Σ}\left[y_i\right]+i \text{ Σ}\left[z_i\right]=0$
$\implies 0\neq 2$
All entries are zero $\therefore |A|=0$